LeetCode 938. Range Sum of BST

原题链接在这里：https://leetcode.com/problems/range-sum-of-bst/

题目：

Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).

The binary search tree is guaranteed to have unique values.

Example 1:

Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
Output: 32

Example 2:

Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
Output: 23

Note:

1. The number of nodes in the tree is at most 10000.
2. The final answer is guaranteed to be less than 2^31.

题解：

If current node is null, return 0.

If node value < L, return range sum from its right.

If node value > R, return range sum from its left.

If it is withing [L, R], return range sum from both left and right + node.val.

Time Complexity: O(n).

Space: O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int rangeSumBST(TreeNode root, int L, int R) {
        if(root == null){
            return 0;
        }
        
        if(root.val < L){
            return rangeSumBST(root.right, L, R);
        }
        
        if(root.val > R){
            return rangeSumBST(root.left, L, R);
        }
        
        return rangeSumBST(root.right, L, R) + rangeSumBST(root.left, L, R) + root.val;
    }
}

AC JavaScript:

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} L
 * @param {number} R
 * @return {number}
 */
var rangeSumBST = function(root, L, R) {
    if(!root){
        return 0;
    }
    
    if(root.val < L){
        return rangeSumBST(root.right, L, R);
    }
    
    if(root.val > R){
        return rangeSumBST(root.left, L, R);
    }
    
    return root.val + rangeSumBST(root.left, L, R) + rangeSumBST(root.right, L, R);
};

Iteration with stack.

Time Complexity: O(n).

Space: O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int rangeSumBST(TreeNode root, int L, int R) {
        if(root == null){
            return 0;
        }
        
        int sum = 0;
        Stack<TreeNode> stk = new Stack<>();
        stk.push(root);

        while(!stk.isEmpty()){
            TreeNode cur = stk.pop();
            if(cur == null){
                continue;
            }
            
            if(cur.val < L){
                stk.push(cur.right);
            }else if(cur.val > R){
                stk.push(cur.left);
            }else{
                stk.push(cur.left);
                stk.push(cur.right);
                sum += cur.val;
            }
        }
        
        return sum;
    }
}